Introduction to integrals Integral, like limit and derivative, is another important
concept in calculus Integral is the inverse of differentiation in some sense There is a connection between integral calculus and
differentiation calculus. The area and distance problems are two typical
applications to introduce the definite integrals
The area problem Problem: find the area of the region S with curved sides, w
hich is bounded by x-axis, x=a, x=b and the curve y=f(x). Idea: first, divide the region S into n subregions by partitio
ning [a,b] into n subintervals [xi-1,xi] (i=1,,n) with x0=a and xn=b; then, approximate each subregion Si by a rectangle since f(x) does not change much and can be treated as a constant in each subinterval [xi-1,xi], that is, Si¼(xi-xi-1)f(i), where i is any point in [xi-1,xi]; last, make sum and take limit if the limit exists, then
the region has area
1n
ii
S1
lim ,
n
in
i
S
1
lim .
n
in
i
S
Remark In the above limit expression, there are two places of signif
icant randomness compared to the normal limit expression: the first is that the nodal points {xi} are arbitrarily chosen, the second is that the sample points {i} are arbitrarily taken too.
means, no matter how {xi} and {i} are
chosen, the limit always exists and has same value.
1
lim
n
in
i
S S
1
lim
n
in
i
S
The distance problem Problem: find the distance traveled by an object during the
time period [a,b], given the velocity function v=v(t). Idea: first, divide the time interval [a,b] into n subintervals;
then, approximate the distance di in each subinterval [ti-1,ti] by di¼(ti-ti-1)v(i) since v(t) does not vary too much and can be treated as a constant; last, make sum and take
limit if the limit exists, then the distance in the
time interval [a,b] is
1n
ii
d
1
lim ,
n
in
i
d
1
lim .
n
in
i
d d
Definition of definite integral We call a partition of the
interval [a,b]. is called the size of the partition,
where are called
sample points. is called Riemann sum. Definition Suppose f is defined on [a,b]. If there exists a
constant I such that for any partition p and any sample points
the Riemann sum has limit then we call
f is integrable on [a,b] and I is the definite integral of f
from a to b, which is denoted by
0 1 1: n np a x x x x b
1max{ }
ii n
x
1( 1, , ). i i ix x x i n 1[ , ]( 1, , ) i i ix x i n
1
( )
n
i ii
f x
i0
1
lim ( ) ,
n
i ii
f x I
( ) .b
aI f x dx
Remark The usual way of partition is the equally-spaced partition
so the size of partition is
In this case is equivalent to
Furthermore, the sample points are usually chosen by
or thus the Riemann sum is given by
, 0,1, ; ( ) /ix a ih i n h b a n
( ) /h b a n
1i ix
0 n
i ix
1
0 1
( ) ( )( ) or ( )
n n
i i
b a i b a b a i b af a f a
n n n n
Example Ex. Determine a region whose area is equal to the given
limit
(1) (2)10
1
2 2lim (5 )
n
ni
i
n n
1
lim tan4 4
n
ni
i
n n
Definition of definite integral In the notation a and b are called the limits of i
ntegration; a is the lower limit and b is the upper limit; f(x) is called the integrand.
The definite integral is a number; it does not depend on x, that is, we can use any letter in place of x:
Ex. Use the definition of definite integral to prove that
is integrable on [a,b], and find ( ) f x c
( ) ,b
af x dx
( )b
af x dx
( ) ( ) ( ) . b b b
a a af x dx f t dt f r dr
.b
acdx
Interpretation of definite integral If the integral is the area under the
curve y=f(x) from a to b If f takes on both positive and negative values, then the int
egral is the net area, that is, the algebraic sum of areas
The distance traveled by an object with velocity v=v(t), during the time period [a,b], is
( )b
af x dx( ) 0,f x
( )b
af x dx
v( ) .b
at dt
Example Ex. Find by definition of definite integral. Sol. To evaluate the definite integral, we partition [0,1] int
o n equally spaced subintervals with the nodal points
Then take as the sample points. By taking limit to the Riemann sum, we have
1 2
0x dx
/ , 0,1, , . ix i n i n / i ix i n
1 2 230
1 1
1 ( 1)(2 1) 1lim ( ) lim ( ) lim .
6 3
n n
i in n n
i i
i n n nx dx f x
n n n
Example Ex. Express the limit into a
definite integral. Sol. Since we have
with Therefore,
The other solution is
1 1 1lim( )
1 2 2
n n n n
1 1 1,
1
in i nn
1 1
1 1 1 1 1 1( ),
1 2 2 1
n n
i i
if
in n n n n nn1
( ) .1
f xx
1
0
1 1 1 1lim( ) .
1 2 2 1
n
dxn n n x
2
1
1. dx
x
Example Ex. If find the limit
Sol.
1 2 ( 1)lim (sin sin sin ).
n
n
n n n n
0
sin 2, xdx
0
1 2 ( 1)lim (sin sin sin )
1 2lim (sin sin sin )
1 2sin .
n
n
n
n n n nn
n n n n
xdx
Exercise1. Express the limits into definite integrals:
(1)
(2)
2. If find
1 2 ( 1)2 2 21
lim (1 ).n
n n n
ne e e
n
2 2 2 2 2 2
2 2 2lim( ).
4 1 4 2 4n n n n n
2sin sin sin
lim( ).1 112
n
nn n n
n n nn
0sin 2, xdx
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